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kvasek [131]
3 years ago
6

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

n does 288 J of work on the surroundings. What is the final volume of the cylinder? Tro, Nivaldo J.. Chemistry (p. 299). Pearson Education. Kindle Edition.
Physics
1 answer:
Step2247 [10]3 years ago
7 0

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

where

p is the pressure of the gas

V_i is the initial volume

V_f is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

V_i = 0.250 L

And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = \frac{28}{10} × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

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Explanation:

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W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

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